Prove that an isolated sigularity of cannot be a pole of .

I know the key is to prove that and have no common pole, but how??

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- February 1st 2010, 07:23 AMynjf and e^f have no common pole?
Prove that an isolated sigularity of cannot be a pole of .

I know the key is to prove that and have no common pole, but how?? - February 1st 2010, 09:11 PMBruno J.
Hello! Here is an idea.

It is easy to see that this is true if is an essential or removable singularity of .

In the case of an -th order pole, then I claim that is an essential singularity of . Clearly it is not a regular point, so it suffices to show that does not tend to infinity as . To do this, it suffices to show that every neighbourhood of contains points such that , for some fixed . - February 1st 2010, 09:38 PMynj
- March 4th 2010, 10:03 PMBruno J.
Hi Ynj!

I know this is a really late answer but I had forgotten about this problem and I remembered it today and solved it. So here it is.

Let have a pole at . I claim that has an essential singularity at .

The problem is solved if we can show that there exists a sequence such that is bounded as ; this is because if were a pole of then we would have as . This is equivalent to finding a sequence such that remains bounded, since . Since we know that necessarily blows up to infinity as as , it is the imaginary part which must become very big.

Consider . This is clearly holomorphic at and . Let be a sequence of bounded open neighbourhoods of whose intersection is . Then is a sequence of neighbourhoods of ; in particular, each of the is bounded, and thus is a sequence of neighbourhoods of infinity. So there are points of the imaginary axis in each of the ; pick such that . Then the sequence converges to and . Thus for and we are done.